showing image from database

You will have to show the content of image, like this[php]echo "<img src =\"" . $myrow['image']."\">"; [/php]Ronald

You will have to show the content of image, like this[php]echo "<img src =\"" . $myrow['image']."\">"; [/php]Ronald

thanks, when i add that line i get the 'X' like the link is gone and then the mime result again, has to be something i am missing???

thanks, when i add that line i get the 'X' like the link is gone and then the mime result again, has to be something i am missing???

Another way is[php]$image = $myrow['image'];
header("Content-type: image/jpeg");
print $image;[/php]Ronald

Another way is[php]$image = $myrow['image'];
header("Content-type: image/jpeg");
print $image;[/php]Ronald

your first way actually worked after i set the image as a link in mysql image field instead of the actual file. think i will leave it like that. Thanks alot.

how about another issue on this subject. how would i get a default image to show up if the person signing is not in my database

I'll put it inside the total code in the ELSE branch:[php]
<?php
include_once('../sql_connect.php');
$result = mysql_query("SELECT A.LastName, A.FirstName, A.Server, A.TimeIn, A.TimeOut, B.image FROM nocsis A, image B WHERE A.LastName = B.LastName AND A.TimeOut IS NULL ORDER BY A.TimeIn ASC");
if (mysql_num_rows($result) > 0) {
while($myrow = mysql_fetch_array($result)) {//begin of loop
//now print the results:
echo "<b><u> Guest:</b></u> ";
echo "<br>Last Name:&nbsp; &nbsp;";
echo $myrow['LastName'];
echo "<br>First Name:&nbsp; &nbsp;";
echo $myrow['FirstName'];
echo "<br>Image:&nbsp; &nbsp;";
echo "<img src =\"" . $myrow['image']."\">";
} //end of loop
} // End IF
else {
echo "<img src ='default.jpg'>";
}
?>
[/php]Ronald

I'll put it inside the total code in the ELSE branch:[php]
<?php
include_once('../sql_connect.php');
$result = mysql_query("SELECT A.LastName, A.FirstName, A.Server, A.TimeIn, A.TimeOut, B.image FROM nocsis A, image B WHERE A.LastName = B.LastName AND A.TimeOut IS NULL ORDER BY A.TimeIn ASC");
if (mysql_num_rows($result) > 0) {
while($myrow = mysql_fetch_array($result)) {//begin of loop
//now print the results:
echo "<b><u> Guest:</b></u> ";
echo "<br>Last Name:&nbsp; &nbsp;";
echo $myrow['LastName'];
echo "<br>First Name:&nbsp; &nbsp;";
echo $myrow['FirstName'];
echo "<br>Image:&nbsp; &nbsp;";
echo "<img src =\"" . $myrow['image']."\">";
} //end of loop
} // End IF
else {
echo "<img src ='default.jpg'>";
}
?>
[/php]Ronald

You have been a great help. appreciate it

hmmm kinda works just no image (i did add a default image) is displayed when the LastName does not match in the table.

also if you're in the mood how would i get the images to show up next to each other rather than in a column

also if you're in the mood how would i get the images to show up next to each other rather than in a column

You might want to check this thread for a neat little work around!

Yes, that is a neat piece of code. I couldn't do it better! Justr replace the url's with the images and their text.

Ronald

You have been a great help. appreciate it

hmmm kinda works just no image (i did add a default image) is displayed when the LastName does not match in the table.

also if you're in the mood how would i get the images to show up next to each other rather than in a column

in the if statement would there be a way to state that if lastname not in table nocsis to display a default image

in the if statement would there be a way to state that if lastname not in table nocsis to display a default image

But that is already in the code, you just have to replace the image name with the path/name of the one you want to display.[php] else {
echo "<img src ='your_image_name'>";[/php]Ronald

But that is already in the code, you just have to replace the image name with the path/name of the one you want to display.[php] else {
echo "<img src ='your_image_name'>";[/php]Ronald

the problem with that is it is searching for a 0 result. when someone signs in whether they have an image or not they still get logged in thus the result not being 0. only time the image appears is when no one is signed in.

So in the ELSE branch you need this statement: Ronald

So in the ELSE branch you need this statement:

Expand|Select|Wrap|Line Numbers

1. "SELECT A.LastName, A.FirstName, A.Server, A.TimeIn, A.TimeOut, FROM nocsis A WHERE A.TimeOut IS NULL ORDER BY A.TimeIn ASC");

Ronald

wouldnt the select statement need to compare the 2 tables somehow and select the LastName in one and not the other

I don't know from which of the two tables you select or what table indicates that he has an image I just wanted to point out that you had to do a new select in case the result of the first select was 0.

Ronald

I don't know from which of the two tables you select or what table indicates that he has an image I just wanted to point out that you had to do a new select in case the result of the first select was 0. .

Ronald

Think I am on the right track. my query works with mysql when i do the else select i get the user that does not have a picture, i just cant seem to figure out how to display it or could it be that my if statement is jacked up. this code is got me going in circles. i didnt think it would be that hard to say if user signs in this table1 and has photo in this table2 display picture from table2 else if user signs in table1 and has no image in table2 show default image. any ideas??